5

(CL0 5.6) Acurve in the plane cquation ol the line MtyGinecoPhnttically by Ive cquations tangjert to the: cutve at Vt and Y 020 Fud2Ifv 10i 11j Zku VxAk Findb.the c...

Question

(CL0 5.6) Acurve in the plane cquation ol the line MtyGinecoPhnttically by Ive cquations tangjert to the: cutve at Vt and Y 020 Fud2Ifv 10i 11j Zku VxAk Findb.the cosine of the angle between ad the vector proj,

(CL0 5.6) Acurve in the plane cquation ol the line MtyGinecoPhnttically by Ive cquations tangjert to the: cutve at Vt and Y 020 Fud 2Ifv 10i 11j Zku Vx Ak Find b.the cosine of the angle between ad the vector proj,



Answers

Vectors Perpendicular to a Plane Find a vector that is perpendicular to the plane passing through the three given points.
$$
P(3,4,5), Q(1,2,3), R(4,7,6)
$$

Now we are given. But Turkey is fight like that minus six point five Jacob better be is three point five vicap minus so on. Jacob, let's see, what you see is CX ICA See why Chacha cx nc Way are its X and Y conference respectively. Now we're also given that a is for particular you see which implies the doctor up off A and C zero. Now we know that dot product is defined us. If we have two vectors and be the doctor, it is defined as the broad, some off the product off there in individual confidence, which would be egg speaks. This a vibe and in this case, a dog would be the components the ex components multiplied, which is five multiplied by C x. Let's divide conference, which is minus six point five C Y is equal to zero. We can rewrite this equation as SI X is equal to six point five by five c y. We simply sent this sea white onto the other side and they were there on both sides by five. Now we also know that the dot product off B and C is fifteen years hence be dot see is fifteen, which gives product off ex components which would be three point five multiplied by c x less. The white confidence which would be minus seven multiplied by C Y, is equal to fifteen. Now we use this equation on eliminated one of the two three variables CX and see why in this equation let a substitute for CX. Now this is cheap on five multiplied way the instead ofthe c x, we can write this expression, which is six point five five five seven. Now we already know this is minus. C Y is equal to fifteen. No, this is a simple mathematical expression. We can do this multiplication on Do the subtraction to find that this number is minus two one four five. See, why is it called a fifteen and we get See why is equal to fifteen by minus fifteen, two point four five, which is Maya's six point one two Now. By substituting this into this equation, we can find c x I'LL see X is equal to six point five by five multiplied by sine a six pound one toe which gives you minus seven point nine six. Hence the vector C is can be written as minus six point one. Toe caps minus seven point nine. I'm sorry. CX is minus seven point nine six, and hence it will be find a seven point nine six minus six point one toe. Jacob, which is the C y. This is our electricity.

In this problem, we are given three points and asked to find or to produce a vector that is perpendicular to the plane that contains those three points. And to do this, we're going to take the vector p Q. And take the cross product of that vector with the vector p R. So how do we get P Q and P. R. Well, to get a peek, you were going to take the point Q. And subtract the point p from it. That's because Q is the end point of the vector and P is the starting point. And so we're going to get one 23 minus three 45 And our factor is going to be the foaming we get negative, too negative, too negative, too. Okay. Now, to get the vector PR, we're going to take the point are and subtract the point p from it. And so we're going to get four seven six, minus three for five. And when we do that, we get the following factor. One, three, one And again, that is equal to the vector. PR for the one of here is equal to the vector p Q. And now we're going to take the cross product of these two doctors by running them in a matrix that has the first row of I, J and K a second row with the components of P Q, which are native to negative to negative two and 1/3 row with the components of PR, which are 13 in one. Okay, and now to take the cross product and get the component for I, we're going to cross out the first column and take the determinant of this two by two matrix when we get negative. Two times one minus native, two times three as native to minus native six United to plus six, which is for then, we always attracts the J component to get back up or not, we're going to cross at the middle column and take the determinant of this to go to Matrix. So we get negative two times one minus negative two times one her native to plus two zero and then to get the K component, we're going to cross out the third calm. You take the determinant of this two by two matrix, which is going to be negative two times three minus data two times one which is negative. Six minus negative to Grenada, six plus two or negative for and we have found our vector. It's going to be four zero negative for.

Were given P Q and R, which are three points. And our task is to find the vector that is perp in the plane, which contains thes three points to start off. We have to find the two vectors p Q and P R. Because these are the vectors that contain the points. B Q and R so P Q and P R starting with P. Q. Where P is our initial point and Q's air terminal point. We can find it by doing zero minus three comma to minus zero. Come on. Negative five minus sirrah, thereby giving this a director equivalent to negatives. Three Two negative five. And we repeat the same process for PR negative two minus three. This time, we're just using the values in our hands to have cute zero minus zero and six minus zero, and now that equals negative five zero six. So these are the two vectors that we have to find the cross product for. So to find the cross product, we're going to use this formula listed out of here except in terms of P Q. NPR. Where you were just going to say it's Peak, U and V. We're gonna say it's PR. So peek you cross. Br is equal to starting from the beginning. You two times V three. So two time six, which is 12 minus you three times V two V two is zero. So that is automatically just gonna be a zero comma. You three times V one. So that's negative. Five times negative five giving us 25 minus. You want times V three. So minus 18. Minus negative. 18. Okay. And then last part. We have you one times V two. So that's gonna be you one times V two negative. Three times zero, which is just zero minus you. Two times V one. So that's two times negative. Five giving us negative 10. Now, when we solve this, we get peak. You cross PR is equal to 12. 43 10 and this is our final answer.

In this problem here, given the three points and told that a plane passes through those points and then asked to produce a vector perpendicular to that plane. And so we're going to do this by taking the vector p Q and taking the cross product of that with the vector PR. Now how do you find P Q and P are defined? Peak. You were going to take the point Q And subtract the point p we're taking Q minus P. Because two is our endpoint and P is our starting point. And so we're going to take zero to negative five and some charge three 00 And the resulting factor is the following negative three to negative five and again that is equal to P Q. And then to get PR willing to take the point arts attract the point p. And when we do that, we get negative to 06 minus 300 And when we do that this we get the resulting sector or should be negative. 50 a six and again, that is he going to PR. I'm going to take the cross product of those two vectors are first up then that is going to be to write a matrix, but its first row being i J K. It's second row being the components of P Q, which is negative. Three to negative five and its third road and components of PR. What you're negative five zero and a six. Okay, now, to take the cross product and get the components of I, we went across out the first call and take the determinants of this two by two matrix just going to be to tempt six months negative five times zero or 12 months, zero just 12. And then we always subtract b j component. And to get that, we're going to cross at the middle, calm and take the come the determinants of this two by two matrix, which is going to be negative three times six minus negative. Five times negative five is going to be negative 18 minus 25 and that is going to be negative 43 and then to notice, cancel out and create a positive. Then to get the components for K, we're going to cross out the third call and take the determines of this two by two matrix, which is going to be negative. Three times zero minus two times negative. Five zero minus negative. 10 or zero plus 10 is just Tim. And we have found our vector. It's going to be 12. 43 okay?


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